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--- 184_notes:examples:week8_wheatstone [2018/06/26 14:48] – curdemma
+++ 184_notes:examples:week8_wheatstone [2021/07/22 18:28] (current) – schram45
@@ Line 19: / Line 19: @@
 ===Approximations & Assumptions===
-  * The wire has very very small resistance when compared to the other resistors in the circuit.
+  * The resistors (including the light bulb) in the circuit are made of Ohmic materials: Ohmic materials have a linear relationship between voltage and current, this allows us to use ohms law.
-  * The circuit is in a steady state.
+  * The wire has very very small resistance when compared to the other resistors in the circuit: This allows there to be no energy loss across the wires and no potential difference across them either simplifying down the model.
-  * Approximating the battery as a mechanical battery.
+  * The circuit is in a steady state: It takes a finite amount of time for a circuit to reach steady state and set up a charge gradient. Making this assumption means the current is not changing with time in any branch of the circuit.
-  * The resistors (including the light bulb) in the circuit are made of Ohmic materials.
+  * Approximating the battery as a mechanical battery: This means the battery will supply a steady power source to the circuit to keep it in steady state.
 ===Representations===
   * We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as
@@ Line 58: / Line 57: @@
 [{{ 184_notes:8_wheatstone_current_directions.png?300 |Circuit with Wheatstone bridge, and directed current}}]
-To find  $\Delta V_{\text{light}}$ , we will need to set up some equations using the Loop Rule and Node Rule. We will focus on Nodes C and D, and the Loops highlighted below. Note that if you want to be able to solve for everything in the circuit, you have to use some loop equations and some node equations (you can't use only loop or only nodes).
+To find  $\Delta V_{\text{light}}$ , we will need to set up some equations using the Loop Rule and Node Rule. We will focus on Nodes A and B, and the Loops highlighted below. Note that if you want to be able to solve for everything in the circuit, you have to use some loop equations and some node equations (you can't use only loop or only nodes).
 [{{ 184_notes:8_wheatstone_loops.png?900 |Highlighted loops in Wheatstone bridge circuit}}]
@@ Line 64: / Line 63: @@
 Applying the Node Rule and the Loop Rule, we obtain the following equations:
 \begin{align*}
-  I &= I_1 + I_2 &(\text{Node C}) \\
+  I &= I_1 + I_2 &(\text{Node A}) \\
-  I &= I_3 + I_4 &(\text{Node D}) \\
+  I &= I_3 + I_4 &(\text{Node B}) \\
   \Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 &(\text{Loop 1}) \\
   \Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 &(\text{Loop 2}) \\
@@ Line 82: / Line 81: @@
 Since we know  $\Delta V_3$  and all the resistances, all that remains is to rearrange and solve for  $\Delta V_{\text{light}}$ :
  $\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$
+Looking at loop 2 alone between the power source and resistor 3 we would expect the voltage across any other elements in that loop to be small. Our answer agrees with this observation as 2.37 is quite small compared to both the battery and resistor 3.